Answer:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is
is the capacitance, 
is the common plate area, 
is the plate separation and 
is the permittivity of the material between the plates.
For air or free space, is 
called the permittivity of free space. In general, 
where 
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, 
.
The energy stored in a capacitor is the average of the product of its charge and voltage.
Its charge, 
, is related to its capacitance by 
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for 
,
A. Substituting for in ,
B. When the distance is 
,
C. When the distance is restored but with a dielectric material of dielectric constant, 
, inserted, we have
Answer:
please the answer below
Explanation:
(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is
k=8.89*10^9
For both cases we have
(b) by replacing this values of r in the expression for V we obtain
hope this helps!!
The inflated balloon shrinks when it is placed in an ice bath with no change in atmospheric pressure.
<u>Explanation:</u>
When the inflated balloon is subjected to an ice bath, it shrinks. This is due to the fact that smaller volume gets occupied by the air/gas inside the balloon as the temperature decreases. Hence, causes the balloon walls to collapse.
An ice bath also lowers the overall air temperature of the balloon inside. As the temperature decreases, the air molecules move more slowly and with lower energy. Because of the particle's lower energy, their collisions with the walls are not enough to keep the inflated balloon.
Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N
